1.

A nucleus X initially at rest, undergoes alpha decay according to the equation `_Z^232Xrarr_90^AY+alpha` What fraction of the total energy released in the decay will be the kinetic energy of the alpha particle?A. (a) `90/92`B. (b) `228/232`C. (c) `sqrt(228/232)`D. (d) `1/2`

Answer» Correct Answer - B
`A=232-4=228`
From conservation of momentum,
`p_alpha=p_gamma=sqrt(2K_alpham_alpha)=sqrt(2K_gammam_gamma)`
or `(K_alpha)/(K_gamma)=(m_gamma)/(m_alpha)=228/4`
`:.` `K_alpha=((228)/(228+4))K_(t otal)`
`=(228/232)K_(T otal)`


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